CHAPTER 27
SOLUTION FOR PROBLEM 23
(a) First find the currents. Let i1 be the current in R1 and take it to be positive if it is to the
right. Let i2 be the current in R2 and take it to be positive if it is to the left. Let i3 be the current
in R3 and take it to be positive if it is to downward. The junction rule produces
i1 + i2 − i3 = 0 .
The loop rule applied to the left-hand loop produces
E1 − i1 R1 − i3 R3 = 0
and applied to the right-hand loop produces
E2 − i2 R2 − i3 R3 = 0 .
Substitute i3 = −i1 + i2 , from the first equation, into the other two to obtain
E1 − i1 R1 − i1 R3 − i2 R3 = 0
and
E2 − i2 R2 − i1 R3 − i2 R3 = 0 .
The first of these yields
i1 =
E1 − i2 R3
.
R1 + R 3
Substitute this into the second equation and solve for i2 . You should obtain
E2 (R1 + R3 ) − E1 R3
R1 R2 + R 2 R3 + R 1 R3
(1.00 V)(4.00 Ω + 5.00 Ω) − (3.00 V)(5.00 Ω)
= −0.158 A .
=
(4.00 Ω)(2.00 Ω) + (2.00 Ω)(5.00 Ω) + (4.00 Ω)(5.00 Ω)
i2 =
Substitute this into the expression for i1 to obtain
i1 =
E1 − i2 R3 3.00 V − (−0.158 A)(5.00 Ω)
= 0.421 A .
=
R1 + R 3
4.00 Ω + 5.00 Ω
Finally,
i3 = i1 + i2 = (0.421 A) + (−0.158 A) = 0.263 A .
Note that the current in R2 is actually to the right. The current in R1 is to the right and the
current in R3 is downward.
(a), (b), and (c) The rate with which energy is dissipated in R1 is
P1 = i21 R1 = (0.421 A)2 (4.00 Ω) = 0.709 W .
The rate with which energy is dissipated in R2 is
P2 = i22 R2 = (−0.158 A)2 (2.00 Ω) = 0.0499 W ,
and the rate with which energy is dissipated in R3 is
P3 = i23 R3 = (0.263 A)2 (5.00 Ω) = 0.346 W ,
(d) and (e) The power of battery 1 is
i1 E1 = (0.421 A)(3.00 V) = 1.26 W
and the power of battery 2 is
i2 E2 = (−0.158 A)(1.00 V) = −0.158 W .
The negative sign indicates that battery 2 is actually absorbing energy from the circuit.
CHAPTER 27
SOLUTION FOR PROBLEM 37
(a), (b), and (c) At t = 0, the capacitor is completely uncharged and the current in the capacitor
branch is as it would be if the capacitor were replaced by a wire. Let i1 be the current in R1 and
take it to be positive if it is to the right. Let i2 be the current in R2 and take it to be positive if it is
downward. Let i3 be the current in R3 and take it to be positive if it is downward. The junction
rule produces i1 = i2 +i3 , the loop rule applied to the left-hand loop produces E −i1 R1 −i2 R2 = 0,
and the loop rule applied to the right-hand loop produces i2 R2 − i3 R3 = 0. Since the resistances
are all the same, you can simplify the mathematics by replacing R1 , R2 , and R3 with R. The
solution to the three simultaneous equations is
i1 =
2(1.2 × 103 V)
2E
=
= 1.1 × 10−3 A
3R 3(0.73 × 106 Ω)
and
i2 = i3 =
1.2 × 103 V
E
=
= 5.5 × 10−4 A .
3R 3(0.73 × 106 Ω)
(d), (e), and (f) At t = ∞, the capacitor is fully charged and the current in the capacitor branch
is zero. Then i1 = i2 and the loop rule yields
E − i1 R1 − i1 R2 = 0 .
The solution is
i1 = i2 =
1.2 × 103 V
E
=
= 8.2 × 10−4 A .
2R 2(0.73 × 106 Ω)
(g) and (h) The potential difference across resistor 2 is V2 = i2 R2 . At t = 0 it is
V2 = (5.5 × 10−4 A)(0.73 × 106 Ω) = 4.0 × 102 V
and at t = ∞ it is
V2 = (8.2 × 10−4 A)(0.73 × 106 Ω) = 6.0 × 102 V .
(i) The graph of V2 versus t is shown to the right.
V2
E/2
E/3
.....................
................................................
............................
..................
...............
.
.
.
.
.
.
.
.
.
.
...
..........
........
........
.......
.......
.
.
.
.
.
......
......
E/6
t
CHAPTER 27
HINT FOR PROBLEM 5
Since the battery is being charged the potential difference across its terminals is given by E + ir,
where E is its emf and r is its internal resistance. Energy is dissipated at a rate that is given
by i2 r and is being converted to chemical energy at a rate of Ei. When the battery is being
discharged the potential difference across its terminals is given by E − ir.
J
ans: (a) 14 V; (b) 1.0 × 102 W; (c) 6.0 × 102 W; (d) 10 V, 1.0 × 102 W
o
CHAPTER 27
HINT FOR PROBLEM 11
R1 and R2 are in parallel and their equivalent resistor is in series with R3 .
J
ans: 4.50 Ω
o
CHAPTER 27
HINT FOR PROBLEM 17
The dissipation rate in R3 is given by i23 R3 , where i3 is the current in R3 . Use the loop and
junction rules to find an expression for i3 in terms of R3 , then set its derivative with respect to
R3 equal to zero and solve for R3 .
J
ans: 1.43 Ω
o
CHAPTER 27
HINT FOR PROBLEM 25
Write one junction and two loop equation, then solve them simultaneously for the current in R3 .
Assume the ammeter has zero resistance.
J
ans: 0.45 A
o
CHAPTER 27
HINT FOR PROBLEM 29
Use the loop equation for the loop containing R1 , R2 , Rs , and Rx . Write an expression for the
potential difference Vb − Va and set it equal to zero. Solve these two equations for Rx .